A

作直径$CF$，连接$BF$，如图，
则$\angle FBC=90^{\circ}$，
$\because \angle BAC+\angle EAD=180^{\circ}$，
而$\angle BAC+\angle BAF=180^{\circ}$，
$\therefore \angle DAE=\angle BAF$，
$\therefore \widehat {DE}=\widehat {BF}$，
$\therefore DE=BF=6$，
$\therefore BC=\sqrt{C{F}^{2}-B{F}^{2}}=8$.
解法二：如图，过点$A$作$AM\bot BC$于$M$，$AN\bot DE$于$N$.

$\because AM\bot BC$，$AN\bot DE$，
$\therefore CM=MB$，$DN=NE=3$，
$\because AC=AB=AD=AE$，
$\therefore \angle BAC=2\angle MAC$，$\angle EAD=2\angle DAN$，
$\because \angle BAC+\angle EAD=180^{\circ}$，
$\therefore 2\angle CAM+2\angle DAN=180^{\circ}$，
$\therefore \angle CAM+\angle DAN=90^{\circ}$，
$\because \angle ACM+\angle CAM=90^{\circ}$，
$\therefore \angle ACM=\angle DAN$，
$\because \angle AMC=\angle AND=90^{\circ}$，
$\therefore \triangle AMC$≌$\triangle DNA\left(AAS\right)$，
$\therefore AM=DN=3$，
$\therefore CM=\sqrt{A{C}^{2}-A{M}^{2}}=\sqrt{{5}^{2}-{3}^{2}}=4$，
$\therefore BC=2CM=8$.
故选：$A$.